Uva 省赛傻逼题 G

太可怕了！

 

 

Problem G

Time limit: 1.000 seconds

 

 

Good Teacher

 

I want to be a good teacher, so at least I need to remember all the student names. However, there are too many students, so I failed. It is a shame, so I don't want my students to know this. Whenever I need to call someone, I call his CLOSEST student instead. For example, there are 10 students:

 

A ? ? D ? ? ? H ? ?

 

Then, to call each student, I use this table:

 

Pos	Reference
1	A
2	right of A
3	left of D
4	D
5	right of D
6	middle of D and H
7	left of H
8	H
9	right of H
10	right of right of H

 

Input

 

There is only one test case. The first line contains n, the number of students (1<=n<=100). The next line contains n space-separated names. Each name is either ? or a string of no more than 3 English letters. There will be at least one name not equal to ?. The next line contains q, the number of queries (1<=q<=100). Then each of the next q lines contains the position p (1<=p<=n) of a student (counting from left).

 

Output

 

Print q lines, each for a student. Note that "middle of X and Y" is only used when X and Y are both closest of the student, and X is always to his left.

 

Sample Input

 

10
A ? ? D ? ? ? H ? ?
4
3
8
6
10

 

Output for the Sample Input

 

left of D
H
middle of D and H
right of right of H

 

---

The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan

 

 

#include <map>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1111
#define ll long long
int n,m;
char s[maxn][11];
int main(){
    while(~scanf("%d",&n)){
        int t;
        for(int i=0;i<n;i++)scanf("%s",s[i]);
        scanf("%d",&t);
        while(t--){
            scanf("%d",&m);
            m--;
            for(int i=0;i<n;i++){
                if(s[m][0]!='?'){printf("%s\n",s[m]);break;}
                if(m-i>=0&&m+i<n&&s[m+i][0]!='?'&&s[m-i][0]!='?'){printf("middle of %s and %s\n",s[m-i],s[m+i]);break;}
                if(m-i>=0&&s[m-i][0]!='?'&&(m+i>=n||s[m+i][0]=='?')){for(int j=0;j<i;j++)printf("right of ");printf("%s\n",s[m-i]);break;}
                if(m+i<n&&s[m+i][0]!='?'&&(m-i<0||s[m-i][0]=='?')){for(int j=0;j<i;j++)printf("left of ");printf("%s\n",s[m+i]);break;}
            }
        }
    }
    return 0;
}